Divide the following complex numbers. $ \dfrac{8+32i}{-5-3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-5+3i}$ $ \dfrac{8+32i}{-5-3i} = \dfrac{8+32i}{-5-3i} \cdot \dfrac{{-5+3i}}{{-5+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(8+32i) \cdot (-5+3i)} {(-5-3i) \cdot (-5+3i)} = \dfrac{(8+32i) \cdot (-5+3i)} {(-5)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(8+32i) \cdot (-5+3i)} {(-5)^2 - (-3i)^2} = $ $ \dfrac{(8+32i) \cdot (-5+3i)} {25 + 9} = $ $ \dfrac{(8+32i) \cdot (-5+3i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({8+32i}) \cdot ({-5+3i})} {34} = $ $ \dfrac{{8} \cdot {(-5)} + {32} \cdot {(-5) i} + {8} \cdot {3 i} + {32} \cdot {3 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-40 - 160i + 24i + 96 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-40 - 160i + 24i - 96} {34} = \dfrac{-136 - 136i} {34} = -4-4i $